Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 121 / 3 / ii

We prove the implication from existence of a witness in the extension to a condition forcing the existential statement. If $M[G]\models \exists x\phi (x)$, choose a witness $a\in M[G]$ and a forcing name $\sigma \in M$ with $\mathrm{val}(\sigma ,G)=a$. The assumed instance of the truth lemma supplies $p\in G$ withThe syntactic forcing relation is monotone under stronger conditions, so every $q\le p$ also forces this same named instance. In particular, the conditions below $p$ having some forced witness are dense below a forcing condition $p$. The existential clause of syntactic forcing therefore givesOnly the truth lemma assumed for individual named instances and the recursive existential clause were used.

For the reverse implication, suppose $p\in G$ and $M\models p{⊩}^{\ast }\exists x\phi (x)$. By the existential clause of syntactic forcing, the setis dense below $p$; the existential quantifier ranges over forcing names of $M$. Definability of the syntactic forcing relation and axiom schema of separation put $D$ in $M$.

To apply genericity, augment it to the globally dense setA condition compatible with $p$ has an extension below $p$, then one in $D$, and an incompatible condition already lies in $D^{\prime }$. Thus $D^{\prime }$ is a dense subset of a forcing order in $M$. The generic filter meets it. Since $p\in G$, directedness prevents $G$ from containing a condition incompatible with $p$, so some $q\in G\cap D$ exists.

There is consequently a name $\sigma \in M$ with $M\models q{⊩}^{\ast }\phi (\sigma )$. The assumed truth lemma gives $M[G]\models \phi (\mathrm{val}(\sigma ,G))$, and henceTogether with part (a), this completes the existential step of the forcing theorem without assuming the desired existential truth lemma in advance.